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A safety valve of 60 mm diameter is to blow off at a pressure of 1.2 N/mm2. .........
Define the following terms with respect to springs...............
Draw a neat sketch of leaf spring of semi-elliptical type and name its parts.
Sketch of Leaf Spring of semi elliptical Type ….Diagram+ Names :
A railway wagon having 1500 kg mass and moving at 1 m/s velocity dashes against a bumper consisting of two helical springs of spring index 6. The springs, which get compressed by 150 mm while resisting a dash made of spring steel having allowable shear st
State any four applications of spring.
1) To cushion, absorb or control energy to external load : Car springs, Railway buffers 2) To store Energy : Watches Toys 3) To Measure forces : Spring Balances, Gauges ,Engines 4) To provide clamping force in Jigs & fixtures. 5) To apply forces as in brakes, clutches & spring loaded valve.
State two applications of leaf spring. Draw neat sketch of leaf spring
Application of Leaf spring Bus/truck/Car suspension springs, diving board, Sketch of Leaf Spring of semi elliptical Type
Given Data: D=250 mm , P=1.5 N/mm2 , n =12 Nos. ,σt = 30 Mpa
Design a helical compression spring with ground ends. The spring index is 12. Maximum load on the spring is 100N and deflection under maximum load is 15 mm. Allowable shear stress of the material is 100 MPa and modulus of rigidity is 4 MPa. Find wire and
Write the equation with Wahl’s factor, used for design of helical coil spring. State the SI unit of each term in the equation
A helical valve spring is to be designed for an operating load range of approximately 135 N. The deflection of the spring for the load range is 7.5 mm. Assume spring index of 10. Permissible shear stress for the material of the spring = 480 MPa and its mo
given load W= 135N Deflection ᵟ =7.5mm Spring index c=10 Permissible shear stress Ʈ=480 MPa Modulus of rigidity G =80 KN/mm2 Wahl’s factor K =4C-1/4C-4 +0.615/C=4X10-1/4X10-4 +0.615/10=1.14 (1)Mean dia. Of the spring coil (1 mark) Maximum shear stress, Ʈ = Kx 8WC/π d 2 480 = 1.14x 8x135x10/3.142xd2 d = 2.857mm from table we shall take a standard wire of size SWG 3 having diameters (d) =2.946mm mean dia. Of the spring coil D= CXd =10x2.946=29.46 mm outer dia.